poj 3041 二分图匹配一般 匈牙利算法

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Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23806 Accepted: 12912

Description

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid. 

Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate all of the asteroids.

Input

* Line 1: Two integers N and K, separated by a single space. 
* Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of times Bessie must shoot.

Sample Input

3 41 11 32 23 2

Sample Output

2

题意：给你长为n的正方形，其中有k个点你要消除，你可以一次性消除一行或者一次性消除一列，问最小消除几次才能全部消除这些点。

思路：这个题想通了就是水题，想不通就gg，其实这个题可以把(x,y)看做对立的点进行二分图匹配，然后，就是模板了。

#include<stdio.h>#include<string.h>int G[500][500];int link[500],used[500];int n,k;int dfs(int u){for(int i=1;i<=n;i++){if(G[u][i]&&!used[i]){used[i]=1;if(!link[i]||dfs(link[i])){link[i]=u;return 1;}}}return 0;}int main(){while(~scanf("%d%d",&n,&k)){int i,u,v,ans=0;memset(G,0,sizeof(G));memset(link,0,sizeof(link));for(i=0;i<k;i++){scanf("%d%d",&u,&v);G[u][v]=1;}for(i=1;i<=n;i++){memset(used,0,sizeof(used));ans+=dfs(i);}printf("%d\n",ans);}}